Repository: coding-minutes/dsa-essentials-solutions-cpp
Branch: master
Commit: 03d2b96551a0
Files: 76
Total size: 53.1 KB

Directory structure:
gitextract_2mb6elmx/

└── DSA Essentials Solutions/
    ├── 2d Arrays/
    │   ├── PascalsTriangle.cpp
    │   ├── SubmatrixSum.cpp
    │   └── WavePrint.cpp
    ├── Arrays/
    │   ├── K-rotate.cpp
    │   ├── LargestElement.cpp
    │   ├── LowerBound.cpp
    │   ├── MaximumSumSubarray.cpp
    │   └── SortedPairSum.cpp
    ├── Backtracking/
    │   ├── N-QueenWays.cpp
    │   ├── RatAndMice.cpp
    │   ├── UniqueSubset.cpp
    │   ├── WordBreakProblem.cpp
    │   └── WordSearch.cpp
    ├── Basic Sorting Algorithms/
    │   ├── Chopsticks.cpp
    │   ├── DefenseKingdom.cpp
    │   ├── OptimisedBubbleSort.cpp
    │   ├── SortingCartesianProducts.cpp
    │   └── SortingWithComparator.cpp
    ├── BinarySearchTree/
    │   ├── DeleteInBST.cpp
    │   ├── IsBST.cpp
    │   └── MirrorABST.cpp
    ├── BinaryTree/
    │   ├── ExpressionTree.cpp
    │   ├── K-thLevel.cpp
    │   ├── MinDepth.cpp
    │   ├── RemoveHalfNodes.cpp
    │   ├── SumOfNodes.cpp
    │   ├── SymmetricTree.cpp
    │   └── TargetPathSum.cpp
    ├── Bit Manipulation/
    │   ├── EarthLevels.cpp
    │   ├── ModuloExponentiation.cpp
    │   ├── SubsetSumQueries.cpp
    │   └── Xoring.cpp
    ├── Divide and Conquer/
    │   ├── 2DArrayMerge.cpp
    │   └── BinarySearchUsingRecursion.cpp
    ├── Dynamic Programming/
    │   ├── CoinChange.cpp
    │   ├── MinimumPartitioning.cpp
    │   ├── OptimalGameStrategy.cpp
    │   └── Vacation.cpp
    ├── Graphs/
    │   ├── AllPathsFromSourceToTarget.cpp
    │   ├── FindStarInGraph.cpp
    │   └── KeysAndRooms.cpp
    ├── Hashing/
    │   ├── ArrayIntersection.cpp
    │   └── KSumSubarray.cpp
    ├── Heaps/
    │   ├── MaximumProduct.cpp
    │   ├── ReduceArraySizeToHalf.cpp
    │   ├── RelativeRanks.cpp
    │   └── WeakestRows.cpp
    ├── Linked List/
    │   ├── AlternateMerge.cpp
    │   ├── BubbleSortOnLinkedList.cpp
    │   ├── DeleteTail.cpp
    │   └── KthLastElement.cpp
    ├── Queues/
    │   ├── FirstNonRepeatingLetter.cpp
    │   ├── InterleaveTwoHalvesOfQueue.cpp
    │   └── SortQueueWithConstantSpace.cpp
    ├── Recursion/
    │   ├── 2DArrayMerge.cpp
    │   ├── AllOccurences.cpp
    │   ├── BinaryStrings.cpp
    │   ├── FriendsParty.cpp
    │   ├── PrintIncreasingNumbers.cpp
    │   └── TilingProblem.cpp
    ├── Stacks/
    │   ├── DuplicateParenthesis.cpp
    │   ├── MaximumRectangularAreaInHistogram.cpp
    │   ├── NextGreaterElement.cpp
    │   ├── ReverseANumberUsingStack.cpp
    │   └── StockSpanProblem.cpp
    ├── Strings/
    │   ├── ArePermutation.cpp
    │   ├── BinaryStringToNumber.cpp
    │   ├── CheckPalindrome.cpp
    │   ├── RemoveDuplicates.cpp
    │   ├── StringCompression.cpp
    │   └── VowelFind.cpp
    ├── Trie/
    │   └── PrefixStrings.cpp
    └── Vectors/
        ├── MakeZeroes.cpp
        ├── RotateImage.cpp
        ├── SortFruits.cpp
        └── SortingCabs.cpp

================================================
FILE CONTENTS
================================================

================================================
FILE: DSA Essentials Solutions/2d Arrays/PascalsTriangle.cpp
================================================
//Expected Time Complexity: O(n^3)

//Hint: Use Binomial Coeffiecients


#include<bits/stdc++.h>
using namespace std;

int binomialCoeff(int n, int k);
 
// Function to print first
// n lines of Pascal's
// Triangle
vector<vector<int>> printPascal(int n)
{
    // Iterate through every line and
    // print entries in it
     vector<vector<int>> res;
    for (int line = 0; line < n; line++)
    {
        // Every line has number of
        // integers equal to line
        // number
vector<int> v;
        for (int i = 0; i <= line; i++)
         {v.push_back(binomialCoeff(line, i));}

         res.push_back(v);
    }
return res;
}
 
int binomialCoeff(int n, int k)
{
    int res = 1;
    if (k > n - k)
    k = n - k;
    for (int i = 0; i < k; ++i)
    {
        res *= (n - i);
        res /= (i + 1);
    }
     
    return res;
}





================================================
FILE: DSA Essentials Solutions/2d Arrays/SubmatrixSum.cpp
================================================
//Hint: Pre Compute Cumilative Sums of every index i,j.

//Expected Time Complexity:
// Pre Computing : O(N^2)
// Queries: O(1)

#include<bits/stdc++.h>
using namespace std;

int sum(vector<vector<int>> v, int sr, int sc, int er, int ec){
    int m=v.size();
    int n=v[0].size();
    // // int aux[m][n];
    int M=m;
    int N=n;
    vector<vector<int>> aux = v;
    vector<vector<int>> mat = v;
    int tli=sr, tlj=sc, rbi=er, rbj=ec;
      for (int i=0; i<N; i++)
      aux[0][i] = mat[0][i];
  
  // Do column wise sum
  for (int i=1; i<M; i++)
      for (int j=0; j<N; j++)
         aux[i][j] = mat[i][j] + aux[i-1][j];
  
  // Do row wise sum
  for (int i=0; i<M; i++)
      for (int j=1; j<N; j++)
         aux[i][j] += aux[i][j-1];
 
 
    int res = aux[rbi][rbj];
  
    // Remove elements between (0, 0) and (tli-1, rbj)
    if (tli > 0)
       res = res - aux[tli-1][rbj];
  
    // Remove elements between (0, 0) and (rbi, tlj-1)
    if (tlj > 0)
       res = res - aux[rbi][tlj-1];
  
    // Add aux[tli-1][tlj-1] as elements between (0, 0)
    // and (tli-1, tlj-1) are subtracted twice
    if (tli > 0 && tlj > 0)
       res = res + aux[tli-1][tlj-1];
  
    return res;
}


================================================
FILE: DSA Essentials Solutions/2d Arrays/WavePrint.cpp
================================================
//Expected Time Complexity: O(n^2)


#include<bits/stdc++.h>
using namespace std;
 
 vector<int> WavePrint(int m, int n, vector<vector<int>> arr)
{
    vector<int> res;
    int i, j = n - 1, wave = 1;
 
    /* m 	- Ending row index
        n 	- Ending column index
        i, j 	- Iterator
        wave 	- for Direction
        wave = 1 - Wave direction down
        wave = 0 - Wave direction up   */
    while (j >= 0) {
         
        // Check whether to go in
        // upward or downward
        if (wave == 1) {
             
            // Print the element of the matrix
            // downward since the value of wave = 1
            for (i = 0; i < m; i++)
                res.push_back(arr[i][j]);           
            wave = 0;
            j--;
        }
        else {
             
            // Print the elements of the
            // matrix upward since the value
            // of wave = 0
            for (i = m - 1; i >= 0; i--)
               res.push_back(arr[i][j]) ;            
            wave = 1;
            j--;
        }
    }
return res;
}


================================================
FILE: DSA Essentials Solutions/Arrays/K-rotate.cpp
================================================
//Expected Time Complexity: O(n)

#include<bits/stdc++.h>
using namespace std;

vector<int> kRotate(vector<int> a, int k)
{
    vector<int> v;
    int n = a.size();
    k = k % n;
 
    for(int i = 0; i < n; i++)
    {
       if(i < k)
       {
           v.push_back(a[n + i - k]);
       }
       else
       {
           v.push_back(a[i - k]);
       }
    }
    return v;
}





================================================
FILE: DSA Essentials Solutions/Arrays/LargestElement.cpp
================================================
//Expected Time Complexity: O(N)

#include<bits/stdc++.h>
using namespace std;

int largestElement(vector<int> A) {
    
     int largestEle = INT_MIN;
     
     for (auto element : A ) {
         largestEle = max(largestEle, element);
     }
     
     return largestEle;
}

================================================
FILE: DSA Essentials Solutions/Arrays/LowerBound.cpp
================================================
//Expected Time Complexity: O(logN)
//Hint: Binary Search


#include<bits/stdc++.h>
using namespace std;


int lowerBound(vector<int> A, int Val) {
    
    int sz = A.size();
    
    int l = 0, r = (sz-1);
    
    int answer = -1;
    
    while (l <= r) {
        int mid = (l + r) / 2;
        if (A[mid] > Val) {
            r = mid-1;
        }
        else {
            answer = A[mid];
            l = mid+1;
        }
    }
    
    return answer;
    
}

================================================
FILE: DSA Essentials Solutions/Arrays/MaximumSumSubarray.cpp
================================================
//Time complexity: O(n)
//Hint: Kadane's Algorithm

#include <bits/stdc++.h>
using namespace std ;

int maxSumSubarray(vector<int> A) {
    int n = A.size(), max_sum = *min_element(A.begin(), A.end()), curr_sum = 0;
    
    for(int i = 0 ; i < n ; i++){
        if(curr_sum + A[i] <= 0){
            curr_sum = A[i];
        }else{
            curr_sum += A[i];
        }
        
        max_sum = max(max_sum, curr_sum);
    }
    
    return max_sum;
}


================================================
FILE: DSA Essentials Solutions/Arrays/SortedPairSum.cpp
================================================
//Expected Time Complexity: O(N)
//Hint: Two Pointer Approach

#include<bits/stdc++.h>
using namespace std;

pair<int, int> closestSum(vector<int> arr, int x){
    int res_l, res_r;  
    int n = arr.size();
    int l = 0, r = n-1, diff = INT_MAX;
 
    while (r > l)
    {
       if (abs(arr[l] + arr[r] - x) < diff)
       {
           res_l = l;
           res_r = r;
           diff = abs(arr[l] + arr[r] - x);
       }
 
       if (arr[l] + arr[r] > x)
           r--;
       else 
           l++;
    }
    
    return {arr[res_l], arr[res_r]};
}

================================================
FILE: DSA Essentials Solutions/Backtracking/N-QueenWays.cpp
================================================
#include <bits/stdc++.h>
using namespace std;


int cnt;

int isSafe(int N, int mat[][20], int r, int c)
{
    // return 0 if two queens share the same column
    for (int i = 0; i < r; i++)
    {
        if (mat[i][c] == 1) {
            return 0;
        }
    }
 
    // return 0 if two queens share the same `` diagonal
    for (int i = r, j = c; i >= 0 && j >= 0; i--, j--)
    {
        if (mat[i][j] == 1) {
            return 0;
        }
    }
 
    // return 0 if two queens share the same `/` diagonal
    for (int i = r, j = c; i >= 0 && j < N; i--, j++)
    {
        if (mat[i][j] == 1) {
            return 0;
        }
    }
 
    return 1;
}

void solve(int N,int mat[][20], int r)
{
    // if `N` queens are placed successfully, print the solution
    if (r == N)
    {
        cnt++;
        return;
    }
 
    // place queen at every square in the current row `r`
    // and recur for each valid movement
    for (int i = 0; i < N; i++)
    {
        // if no two queens threaten each other
        if (isSafe(N, mat, r, i))
        {
            // place queen on the current square
            mat[r][i] = 1;
 
            // recur for the next row
            solve(N,mat, r + 1);
 
            // backtrack and remove the queen from the current square
            mat[r][i] = 0;
        }
    }
}
 

int nQueen(int n){
    cnt =0;
    int arr[20][20]={0};

    solve(n,arr,0);
    return cnt;

}

================================================
FILE: DSA Essentials Solutions/Backtracking/RatAndMice.cpp
================================================
#include<bits/stdc++.h>
using namespace std;

void ratchase(vector<string> a,vector<vector<int>> &b,vector<vector<int>> &v,int i,int j,int n,int m){
	if(i==n && j==m){
		for(int k=0;k<=n;k++){
			for(int l=0;l<=m;l++){
				v[k][l] = b[k][l];
			}cout<<endl;
		}
		return;
	}
	if(i!=n && a[i+1][j]!='X' && b[i+1][j]!=1){
		b[i+1][j]=1;
		ratchase(a,b,v,i+1,j,n,m);
		b[i+1][j]=0;
	}
	if(i>0 && a[i-1][j]!='X' && b[i-1][j]!=1){
		b[i-1][j]=1;
		ratchase(a,b,v,i-1,j,n,m);
		b[i-1][j]=0;
	}
	if(j!=m && a[i][j+1]!='X' && b[i][j+1]!=1){
		b[i][j+1]=1;
		ratchase(a,b,v,i,j+1,n,m);
		b[i][j+1]=0;
	}
	if(j>0 && a[i][j-1]!='X' && b[i][j-1]!=1){
		b[i][j-1]=1;
		ratchase(a,b,v,i,j-1,n,m);
		b[i][j-1]=0;
	}
	return;
}
vector<vector<int>> ratAndMice(vector<string> a) {
    int n = a.size();
    int m = a[0].size();
    vector<vector<int>> v(n, vector<int>(m, 0));
    vector<vector<int>> b(n, vector<int>(m, 0));
    b[0][0] = 1;
    
	ratchase(a,b,v,0,0,n-1,m-1);
	return v;
}

================================================
FILE: DSA Essentials Solutions/Backtracking/UniqueSubset.cpp
================================================
#include <bits/stdc++.h>
using namespace std;
set<vector<int>> s;
void recur(vector<int> &nums, vector<int> ans, int i)
{
    if(i == nums.size()){
        sort(ans.begin(), ans.end());
        s.insert(ans);
        return;
    }
    
    ans.push_back(nums[i]);
    recur(nums, ans, i+1);
    ans.pop_back();
    recur(nums, ans, i+1);
}
vector<vector<int>> uniqueSubsets(vector<int> nums){
    s.clear();
    vector<int> ans;
    recur(nums, ans, 0);
    vector<vector<int>> v;
    for(auto x : s) 
        v.push_back(x);
    return v;
}

================================================
FILE: DSA Essentials Solutions/Backtracking/WordBreakProblem.cpp
================================================
#include<bits/stdc++.h>
using namespace std;

int cnt = 0;
vector<string> v;
void help(string s, int n, string res, vector<string> &word)
{
	for (int i = 1; i <= n; i++)
	{
		string ss = s.substr(0, i);
		int l = word.size();
		bool flag = false;

		for (int j = 0; j < l; j++)
			if (word[j] == ss)
				flag = true;

		if (flag)
		{
			if (i == n)
			{
				res += ss;
				// v.push_back(res);
				cnt++;
				return;
			}
			help(s.substr(i, n - i), n - i, res + ss + " ", word);
		}
	}
}

int wordBreak(string s, vector<string> &dictionary)
{
	cnt = 0;
	// v.clear();
	help(s, s.size(), "", dictionary);
	// for (auto x : v) cout << x << '\n';
	return cnt;
}

================================================
FILE: DSA Essentials Solutions/Backtracking/WordSearch.cpp
================================================
#include <bits/stdc++.h>
using namespace std;
bool chk;
void recur(vector<vector<char>>& board, string &word, int i, int j, int k)
{
    if(k == word.size())
    {
        chk = true;
        return;
    }
    if(i < 0 || j < 0 || i == board.size() || j == board[0].size() || board[i][j] == '-1')
        return;
    
    if(board[i][j] == word[k])
    {
        char c = board[i][j];
        board[i][j] = '-1';
        recur(board, word, i, j+1, k+1);
        recur(board, word, i+1, j, k+1);
        recur(board, word, i, j-1, k+1);
        recur(board, word, i-1, j, k+1);
        board[i][j] = c;
    }
    return;
}
bool wordSearch(vector<vector<char>> &board, string word)
{
    chk  = false;
    for(int i=0; i<board.size(); i++)
    {
        for(int j=0; j<board[0].size(); j++)
        {
            if(board[i][j]==word[0])
                recur(board, word, i, j, 0);
        }
    }
    
    return chk;
}

================================================
FILE: DSA Essentials Solutions/Basic Sorting Algorithms/Chopsticks.cpp
================================================
//Expected Complexity: O(N logN)


#include<bits/stdc++.h>
using namespace std;

int pairSticks(vector<int> length, int D)
{
    // your code goes here
    sort(length.begin(), length.end());
    int res = 0;

    for(int i=0; i<length.size()-1; i++)
    {
        if (length[i + 1] - length[i] <= D) { res++; i++;}
    }

    return res;
}

================================================
FILE: DSA Essentials Solutions/Basic Sorting Algorithms/DefenseKingdom.cpp
================================================
//Expected Time Complexity= O(N log N)
//Find the largest distance from both W and H axis and take their product.



#include<bits/stdc++.h>
using namespace std;
using ll = long long;
#define all(v) v.begin(), v.end()
int defkin(int W, int H, vector<pair<int, int>> position)
{
    // your code goes here
    vector<pair<int, int>> v = position;
    int w = W, h = H;
    vector<ll> x, y;
    x.push_back(0); y.push_back(0);
    
    ll maxx = INT_MIN, maxy = INT_MIN;

    for(int i=0; i<v.size(); i++)
    {
        x.push_back(v[i].first), y.push_back(v[i].second);
    }
    // x.push_back(W); y.push_back(H);
    sort(all(x));
    sort(all(y));

    for (ll i = 1; i < x.size(); i++) maxx = max(maxx, x[i] - x[i - 1] - 1);
    for (ll i = 1; i < y.size(); i++) maxy = max(maxy, y[i] - y[i - 1] - 1);
    maxx = max(maxx, W - x[x.size() - 1] );
    maxy = max(maxy, H- y[y.size() - 1]);

    return (maxx * maxy);
    
}

================================================
FILE: DSA Essentials Solutions/Basic Sorting Algorithms/OptimisedBubbleSort.cpp
================================================
#include<bits/stdc++.h>
using namespace std;

vector<int> optimizedBubbleSort(vector<int> arr){
    // your code  goes here
    int i, j, n=arr.size();
   bool swapped;
   for (i = 0; i < n-1; i++)
   {
     swapped = false;
     for (j = 0; j < n-i-1; j++)
     {
        if (arr[j] > arr[j+1])
        {
           swap(arr[j], arr[j+1]);
           swapped = true;
        }
     }
 
     // IF no two elements were swapped by inner loop, then break
     if (swapped == false)
        break;
   }
   
   return arr;
}

================================================
FILE: DSA Essentials Solutions/Basic Sorting Algorithms/SortingCartesianProducts.cpp
================================================
//Expected Time Complexity :O(N log N)


#include<bits/stdc++.h>
using namespace std;

vector<pair<int, int>> sortCartesian(vector<pair<int, int>> v)
{
    sort(v.begin(), v.end());
    return v;
}

================================================
FILE: DSA Essentials Solutions/Basic Sorting Algorithms/SortingWithComparator.cpp
================================================
#include<bits/stdc++.h>
using namespace std;
bool compare(int a, int b){
    return a > b;
}
vector<int> sortingWithComparator(vector<int> v, bool flag){
    // your code  goes here
    if(flag) sort(v.begin(), v.end());
    else sort(v.begin(), v.end(), compare);
    return v;
}

================================================
FILE: DSA Essentials Solutions/BinarySearchTree/DeleteInBST.cpp
================================================
#include <iostream>
using namespace std;
class node{
    public:
     int data;
     node*left;
     node*right;
     node(int d){
       data=d;
       left=NULL;
       right=NULL;
     }
 };
 
 
 node* deleteNode(node* root, int k){
    // Base case
    if (root == NULL)
        return root;
 
    // Recursive calls for ancestors of
    // node to be deleted
    if (root->data > k) {
        root->left = deleteNode(root->left, k);
        return root;
    }
    else if (root->data < k) {
        root->right = deleteNode(root->right, k);
        return root;
    }
 
    // We reach here when root is the node
    // to be deleted.
 
    // If one of the children is empty
    if (root->left == NULL) {
        node* temp = root->right;
        delete root;
        return temp;
    }
    else if (root->right == NULL) {
        node* temp = root->left;
        delete root;
        return temp;
    }
 
    // If both children exist
    else {
 
        node* succParent = root;
 
        // Find successor
        node* succ = root->right;
        while (succ->left != NULL) {
            succParent = succ;
            succ = succ->left;
        }
 
        // Delete successor.  Since successor
        // is always left child of its parent
        // we can safely make successor's right
        // right child as left of its parent.
        // If there is no succ, then assign
        // succ->right to succParent->right
        if (succParent != root)
            succParent->left = succ->right;
        else
            succParent->right = succ->right;
 
        // Copy Successor Data to root
        root->data = succ->data;
 
        // Delete Successor and return root
        delete succ;
        return root;
    }
}


================================================
FILE: DSA Essentials Solutions/BinarySearchTree/IsBST.cpp
================================================
#include<bits/stdc++.h>
using namespace std;

class Node
{
  public:
   int key;
   Node *left;
   Node *right;

   Node(int key){
       this->key = key;
       left = right  = NULL;
   }
};

bool isBSTUtil(Node* node, int min, int max)
{
    /* an empty tree is BST */
    if (node==NULL)
        return true;
             
    /* false if this node violates
    the min/max constraint */
    if (node->key < min || node->key > max)
        return false;
     
    /* otherwise check the subtrees recursively,
    tightening the min or max constraint */
    return
        isBSTUtil(node->left, min, node->key) &&
        isBSTUtil(node->right, node->key, max); 
}

bool isBST(Node * root){
    //complete this method
    return isBSTUtil(root, INT_MIN,INT_MAX);
    
}


================================================
FILE: DSA Essentials Solutions/BinarySearchTree/MirrorABST.cpp
================================================
#include<bits/stdc++.h>
using namespace std;

class Node
{
  public:
   int key;
   Node *left;
   Node *right;

   Node(int key){
       this->key = key;
       left = right  = NULL;
   }
};

void mirror(Node* node)
{
    if (node == NULL)
        return;
    else
    {
        struct Node* temp;
         
        /* do the subtrees */
        mirror(node->left);
        mirror(node->right);
     
        /* swap the pointers in this node */
        temp     = node->left;
        node->left = node->right;
        node->right = temp;
    }
}

Node* mirrorBST(Node * root){
    //complete this method
    mirror(root);
    return root;
    
}


================================================
FILE: DSA Essentials Solutions/BinaryTree/ExpressionTree.cpp
================================================
//Expected Time Complexity: O(n)
  

#include <bits/stdc++.h>
using namespace std;
 
struct Node {
    string key;
    Node* left, *right;
};

bool isOp(string data)
{
    if(data == "+" or data == "-" or data == "*" or data == "/")
        return true;
    return false;
}

int evalTree(Node* root){
    if(root == NULL) return 0;
    if(!isOp(root->key)) return stoi(root->key);
    
    if(root->key == "+") return evalTree(root->left)+evalTree(root->right);
    if(root->key == "-") return evalTree(root->left)-evalTree(root->right);
    if(root->key == "*") return evalTree(root->left)*evalTree(root->right);
    if(root->key == "/") return evalTree(root->left)/evalTree(root->right);
}

================================================
FILE: DSA Essentials Solutions/BinaryTree/K-thLevel.cpp
================================================
//Expected Time Complexity: O(n)

#include <bits/stdc++.h>
using namespace std;
 
struct Node {
    int key;
    Node* left, *right;
};

vector<int> printKthLevel(Node* root, int k){
    // your code goes here
    // Create Queue
    queue<struct Node*> que;
 
    // Enqueue the root node
    que.push(root);
 
    // Create a set
    vector<int> s;
 
    // Level to track
    // the current level
    int level = 0;
    int flag = 0;
 
    // Iterate the queue till its not empty
    while (!que.empty()) {
 
        // Calculate the number of nodes
        // in the current level
        int size = que.size();
 
        // Process each node of the current
        // level and enqueue their left
        // and right child to the queue
        while (size--) {
            struct Node* ptr = que.front();
            que.pop();
 
            // If the current level matches the
            // required level then add into set
            if (level == k) {
 
                // Flag initialized to 1
                flag = 1;
 
                // Inserting node data in set
                s.push_back(ptr->key);
            }
            else {
 
                // Traverse to the left child
                if (ptr->left)
                    que.push(ptr->left);
 
                // Traverse to the right child
                if (ptr->right)
                    que.push(ptr->right);
            }
        }
 
        // Increment the variable level
        // by 1 for each level
        level++;
 
        // Break out from the loop
        // if the Kth Level is reached
        if (flag == 1)
            break;
    }
    return s;
}

================================================
FILE: DSA Essentials Solutions/BinaryTree/MinDepth.cpp
================================================
#include <bits/stdc++.h>
using namespace std;
 
struct Node {
    int key;
    Node* left, *right;
};

int minDepth(Node *root) {
        // Your code here
        int res = INT_MAX;
        queue<pair<Node*, int>> q;
        int d = 1;
        q.push({root, d});
        while(!q.empty())
        {
            Node* f = q.front().first;
            d = 1+q.front().second;
            if(f->left == NULL && f->right == NULL) 
                res = min(res, q.front().second);
            q.pop();
            if(f->left) q.push({f->left, d});
            if(f->right) q.push({f->right, d});
        }
        return res;
    }

================================================
FILE: DSA Essentials Solutions/BinaryTree/RemoveHalfNodes.cpp
================================================
#include <bits/stdc++.h>
using namespace std;
 
struct Node {
    int key;
    Node* left, *right;
};

void inorder(Node* root, vector<int> &v)
{
    if(root == NULL) return;
    if(root->left) inorder(root->left, v);
    v.push_back(root->key);
    if(root->right) inorder(root->right, v);
}

Node *help(Node *root)
{
   //add code here.
   if(root==NULL) return NULL;
   if(root->right) root->right = help(root->right);
   if(root->left) root->left = help(root->left);
   if( (root->left != NULL && root->right == NULL) or (root->left == NULL && root->right !=NULL) )
   {
       if(root->left) root = root->left;
       else root = root->right;
       root = help(root);
   }
   return root;
   
}

vector<int> removeHalfNodes(Node *root)
{
   //add code here.
   root = help(root);
   vector<int> v;
   inorder(root, v);
   return v;
   
}

================================================
FILE: DSA Essentials Solutions/BinaryTree/SumOfNodes.cpp
================================================
#include <bits/stdc++.h>
using namespace std;
 
struct Node {
    int key;
    Node* left, *right;
};

int sumBT(Node* root)
{
    // Code here
    int res = 0;
    queue<Node*> q;
    q.push(root);
    while(!q.empty())
    {
        Node* f = q.front();
        q.pop();
        if(f->left) q.push(f->left);
        if(f->right) q.push(f->right);
        res += f->key;
    }
    return res;
}

================================================
FILE: DSA Essentials Solutions/BinaryTree/SymmetricTree.cpp
================================================
#include <bits/stdc++.h>
using namespace std;
 
struct Node {
    int key;
    Node* left, *right;
};
bool isSymmetric(Node* root) {
        queue<Node*> q1;
        queue<Node*> q2;
        if(root == NULL || (root->right==NULL && root->left==NULL)) return true;
        if(root->right == NULL || root->left == NULL) return false;
        q1.push(root->left);
        q2.push(root->right);
        while(!q1.empty() && !q2.empty())
        {
            Node* f1 = q1.front();
            q1.pop();
            Node* f2 = q2.front();
            q2.pop();
            if(q1.empty() && !q2.empty()) return false;
            if(!q1.empty() && q2.empty()) return false;
            if(f1->left==NULL && f2->right!=NULL) return false;
            if(f1->left!=NULL && f2->right==NULL) return false;
            if(f1->key != f2->key) return false;
            if(f1->left) q1.push(f1->left);
            if(f1->right) q1.push(f1->right);
            if(f2->right) q2.push(f2->right);
            if(f2->left) q2.push(f2->left);
        }
        if(q1.empty() && q2.empty()) return true;
        return false;
    }

================================================
FILE: DSA Essentials Solutions/BinaryTree/TargetPathSum.cpp
================================================
#include <bits/stdc++.h>
using namespace std;
 
struct Node {
    int val;
    Node* left, *right;
};

vector<vector<int>> vv;
void help(Node* root, int a, vector<int> &v, int b)
{
    if(root == NULL) return;
    if(root->left == NULL && root->right == NULL)
    {
        
        if(a == b+root->val) 
        {
            v.push_back(root->val);
            vv.push_back(v);
            v.pop_back();
        }
        return;
    }
    if(root->left)
    {
        v.push_back(root->val);
        help(root->left, a, v, b+root->val);
        v.pop_back();
    }
    if(root->right)
    {
        v.push_back(root->val);
        help(root->right, a, v, b+root->val);
        v.pop_back();
    }
}
vector<vector<int>> pathSum(Node* root, int targetSum) {
    vv.clear();
    vector<int> v;
    help(root, targetSum, v, 0);
    return vv;
}

================================================
FILE: DSA Essentials Solutions/Bit Manipulation/EarthLevels.cpp
================================================
// Expected Time Complexity : O(Log n)


#include<bits/stdc++.h>
using namespace std;

long long convertDecimalToBinary(unsigned long long int n)
{
    long long binaryNumber = 0;
    unsigned long long int remainder, i = 1, step = 1;

    while (n!=0)
    {
        remainder = n%2;
        n /= 2;
        binaryNumber += remainder*i;
        i *= 10;
    }
    return binaryNumber;
}

int earthLevel(int k)
{
    //your code goes here
    unsigned long long int binaryNumber, sum = 0;
    binaryNumber = convertDecimalToBinary(k);
    
    while (binaryNumber != 0)
    {
        unsigned long long int t;
        t = binaryNumber%2;
        sum = sum + t;
        binaryNumber = binaryNumber/10;
    }
    
    return sum;
}

================================================
FILE: DSA Essentials Solutions/Bit Manipulation/ModuloExponentiation.cpp
================================================
// Expected Time Complexity : O(Log N)
// Hint: if a is even, than x^a can be written as (x^(a/2))*(x^(a/2))

#include <iostream>
using namespace std;
 
 
int power(int x, int y, int p)
{
    int res = 1;     // Initialize result
 
    x = x % p; // Update x if it is more than or
                // equal to p
  
    if (x == 0) return 0; // In case x is divisible by p;
 
    while (y > 0)
    {
        // If y is odd, multiply x with result
        if (y & 1)
            res = (res*x) % p;
 
        // y must be even now
        y = y>>1; // y = y/2
        x = (x*x) % p;
    }
    return res;
}
 

================================================
FILE: DSA Essentials Solutions/Bit Manipulation/SubsetSumQueries.cpp
================================================
// Expected Time Complexity : O(1) for each query


#include<bits/stdc++.h>
using namespace std;

vector<bool> subsetSum(vector<int> v, vector<int> q)
{
    int n = q.size();
    vector<bool> b(n);
    
    bitset<10000> bit;
    bit.reset();
    bit[0] = 1;
  
    for (int i = 0; i < v.size(); ++i)
        bit |= (bit << v[i]);
        
    for(int i=0; i<n; i++)
    {
        int x = q[i];
        bit[x]? b[i]=true : b[i]=false;
    }
    
    return b;
}

================================================
FILE: DSA Essentials Solutions/Bit Manipulation/Xoring.cpp
================================================
// Expected Time Complexity : O(N)
// xor of two same elements is 0

#include<bits/stdc++.h>
using namespace std;

int xoring(vector<int> v)
{
    int res=0;
    for(auto x : v)  res ^= x;
    return res;
}

================================================
FILE: DSA Essentials Solutions/Divide and Conquer/2DArrayMerge.cpp
================================================

#include<bits/stdc++.h>
using namespace std;

void merge_row(vector<vector<int>> &mat,int i, int cs, int cm, int ce){
   vector<int> sorted;
   int x=cs;
   int y=cm+1;
   //cout<<x<<" "<<cm<<" "<<y<<" "<<ce<<endl;
   while(x<=cm && y<=ce){
       if(mat[i][x]<mat[i][y]){
           sorted.push_back(mat[i][x]);
           x++;
       }
       else{
           sorted.push_back(mat[i][y]);
           y++;
       }
   }
   
   
   while(x<=cm){
       sorted.push_back(mat[i][x]);
       x++;
   }
   while(y<=ce){
       sorted.push_back(mat[i][y]);
       y++;
   }
   int k=0;
   for(int j=cs; j<=ce; j++){
       mat[i][j]=sorted[k];
       k++;
   }
   return;
}
void merge_col(vector<vector<int>> &mat,int j, int rs, int rm, int re){
       vector<int> sorted;
   int x=rs;
   int y=rm+1;
   while(x<=rm && y<=re){
       if(mat[x][j]<mat[y][j]){
           sorted.push_back(mat[x][j]);
           x++;
       }
       else{
           sorted.push_back(mat[y][j]);
           y++;
       }
   }
   while(x<=rm){
       sorted.push_back(mat[x][j]);
       x++;
   }
   while(y<=re){
       sorted.push_back(mat[y][j]);
       y++;
   }
   int k=0;
   for(int i=rs; i<=re; i++){
       mat[i][j]=sorted[k];
       k++;
   }
   return;
}


void merge(int m, int n, vector<vector<int>> &mat, int rs, int rm, int re,int cs, int cm, int ce){
    
    
    
    //for sorting rows
    for(int i=rs; i<=re; i++){
        merge_row(mat,i,cs,cm,ce);
    }
    
    //for sorting columns
    for(int j=cs; j<=ce; j++){
        merge_col(mat,j,rs,rm,re);
    }
    return;

}

void merge_sort(int m, int n, vector<vector<int>> &mat, int rs, int re, int cs, int ce){
    //cout<<rs<<" "<<re<<endl;
    //cout<<cs<<" "<<ce<<endl;
     if(rs>=re && cs>=ce){
         return;
     }


     int rm=(rs+re)/2;
     int cm=(cs+ce)/2;
      
    // cout<<rs<<" "<<rm<<" "<<re<<" "<<cs<<" "<<cm<<" "<<ce<<endl; 
     
     
     //for dividing into subarrays
     merge_sort(m,n,mat,rs,rm,cs,cm);
     merge_sort(m,n,mat,rm+1,re,cs,cm);
     merge_sort(m,n,mat,rs,rm,cm+1,ce);
     merge_sort(m,n,mat,rm+1,re,cm+1,ce);

     
    //for merging sorted subarrays
     merge(m,n,mat,rs,rm,re,cs,cm,ce);
     return;
}

vector<vector<int>> mergeSort(int m, int n, vector<vector<int>> v){
    merge_sort(m,n,v,0,m-1,0,n-1);
    return v;
}


================================================
FILE: DSA Essentials Solutions/Divide and Conquer/BinarySearchUsingRecursion.cpp
================================================
//Expected Time Complexity: O(logn)

#include <bits/stdc++.h>
using namespace std;
  
#include <bits/stdc++.h>
using namespace std;
  
// A recursive binary search function. It returns
// location of x in given array arr[l..r] is present,
// otherwise -1
int binary(vector<int> arr, int l, int r, int x)
{
    if (r >= l) {
        int mid = l + (r - l) / 2;
  
        // If the element is present at the middle
        // itself
        if (arr[mid] == x)
            return mid;
  
        // If element is smaller than mid, then
        // it can only be present in left subarray
        if (arr[mid] > x)
            return binary(arr, l, mid - 1, x);
  
        // Else the element can only be present
        // in right subarray
        return binary(arr, mid + 1, r, x);
    }
  
    // We reach here when element is not
    // present in array
    return -1;
}

int binarySearch(vector<int> v, int x)
{
    // your code goes here
    
    int n = v.size();
    int result = binary(v, 0, n - 1, x);
    return result;
}

================================================
FILE: DSA Essentials Solutions/Dynamic Programming/CoinChange.cpp
================================================
#include<bits/stdc++.h>
using namespace std;

long long coinChange(int s, int n , vector<int> a, long long dp[500][100])
{
    if (n < 0 || s < 0) return 0;
    if (s == 0) return 1;

    if (dp[s][n] != 0) return dp[s][n];
    
    long long op1 = coinChange(s, n - 1, a, dp);
    long long op2 = coinChange(s - a[n], n, a, dp);

    return dp[s][n] = op1 + op2;
}

long long findCombinations(int n, vector<int> coins)
{
    long long dp[500][100] = {{0}};
    return coinChange(n, coins.size()-1, coins, dp);
}

================================================
FILE: DSA Essentials Solutions/Dynamic Programming/MinimumPartitioning.cpp
================================================
#include <bits/stdc++.h>
using namespace std;
  
int findMin(vector<int> arr)
{
    int n = arr.size();
    int sum = 0;
    for (int i = 0; i < n; i++)
        sum += arr[i];
  
    // Create an array to store results of subproblems
    bool dp[n + 1][sum + 1];
  
    // Initialize first column as true. 0 sum is possible
    // with all elements.
    for (int i = 0; i <= n; i++)
        dp[i][0] = true;
  
    // Initialize top row, except dp[0][0], as false. With
    // 0 elements, no other sum except 0 is possible
    for (int i = 1; i <= sum; i++)
        dp[0][i] = false;
  
    // Fill the partition table in bottom up manner
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= sum; j++) {
            // If i'th element is excluded
            dp[i][j] = dp[i - 1][j];
  
            // If i'th element is included
            if (arr[i - 1] <= j)
                dp[i][j] |= dp[i - 1][j - arr[i - 1]];
        }
    }
  
    // Initialize difference of two sums.
    int diff = INT_MAX;
  
    // Find the largest j such that dp[n][j]
    // is true where j loops from sum/2 t0 0
    for (int j = sum / 2; j >= 0; j--) {
        // Find the
        if (dp[n][j] == true) {
            diff = sum - 2 * j;
            break;
        }
    }
    return diff;
}

================================================
FILE: DSA Essentials Solutions/Dynamic Programming/OptimalGameStrategy.cpp
================================================
#include <bits/stdc++.h>
using namespace std;

int game(int n, vector<int> v, int s, int e){

    if(s==e || s==e-1){
        return max(v[s],v[e]);
    }

    int op1=v[s] + min(game(n,v,s+2,e),game(n,v,s+1,e-1));
    int op2=v[e] + min(game(n,v,s+1,e-1),game(n,v,s,e-2));
    return max(op1,op2); 
}

int MaxValue(int n, vector<int> v){
    int res=game(n,v,0,n-1);
    return res;
}

================================================
FILE: DSA Essentials Solutions/Dynamic Programming/Vacation.cpp
================================================
#include<bits/stdc++.h>
using namespace std;

using ll = long long;
using vi = vector<int>;
using vl = vector<ll>;
using vvl = vector<vl>;
using vs = vector<string>;
using mi = map<int, int>;
using ml = map<ll, ll>;
using umi = unordered_map<int, int>;
using uml = unordered_map<ll, ll>;
using pi = pair<int, int>;
using pl = pair<ll, ll>;
using ti = tuple<int, int, int>;
using tl = tuple<ll, ll, ll>;
using vii = vector<pi>;
using viii = vector<ti>;
using vll = vector<pl>;
using vlll = vector<tl>;
#define mem(dp) memset(dp, -1, sizeof(dp))
#define aut(a, b) for (auto&(a) : (b))
#define out(x, v) for (auto&(x) : (v)) cout << x << " "; cout << '\n';
#define rep(i, n) for (ll (i) = 0; (i) < (n); ++(i) )
#define repp(i, n) for (ll (i) = 1; (i) <= (n); ++(i) )
#define all(v) v.begin(), v.end()
#define fi get<0>
#define se get<1>
#define th get<2>
#define F first
#define S second
#define mp make_pair
#define mt make_tuple
#define pb push_back
int topDown(viii &v, int n, int i, int dp[][4], int prev)
{
    if (i == n) return 0;
    if (dp[i][prev] != -1) return dp[i][prev];

    int op1 = INT_MIN, op2 = INT_MIN, op3 = INT_MIN;

    if (prev != 1) op1 = fi(v[i]) + topDown(v, n, i + 1, dp, 1);
    if (prev != 2) op2 = se(v[i]) + topDown(v, n, i + 1, dp, 2);
    if (prev != 3) op3 = th(v[i]) + topDown(v, n, i + 1, dp, 3);

    return dp[i][prev] = max(op1, max(op2, op3));
}
int vacation(vector<int> a, vector<int> b, vector<int> c)
{
    viii v;
    int n = a.size();
    rep(i, n)
    {
        v.pb(mt(a[i], b[i], c[i]));
    }
    int dp[n][4];
    mem(dp);
    return topDown(v, n, 0, dp, 0);
}

================================================
FILE: DSA Essentials Solutions/Graphs/AllPathsFromSourceToTarget.cpp
================================================
#include<bits/stdc++.h>
using namespace std;
void dfs(vector<vector<int>>& graph, vector<vector<int>>& result, vector<int> path, int src, int dst) {
    path.push_back(src);
    if(src == dst) {
        result.push_back(path);
        return;
    }

    for(auto node : graph[src])
        dfs(graph, result, path, node, dst);
}
vector<vector<int>> allPathsSourceTarget(vector<vector<int>>& graph) {
    vector<vector<int>> paths; vector<int> path;
    int nodes = graph.size();
    if(nodes == 0) return paths;
    dfs(graph, paths, path, 0, nodes - 1);
    return paths;
}



================================================
FILE: DSA Essentials Solutions/Graphs/FindStarInGraph.cpp
================================================
#include<bits/stdc++.h>
using namespace std;

int findCenter(vector<vector<int>>& v) {
            
           pair<int, int> f={v[0][0],v[0][1]};
           pair<int, int> s={v[1][0],v[1][1]};
        
        if(f.first==s.first){
            return f.first;
        }
        else if(f. first==s.second){
            return f.first;
        }
        else if(f.second==s.first){
            return f.second;
        }
        else{
            return f.second;
        }
        
        
        
        
    }


================================================
FILE: DSA Essentials Solutions/Graphs/KeysAndRooms.cpp
================================================
#include<bits/stdc++.h>
using namespace std;

bool canVisitAllRooms(vector<vector<int>>& rooms) {
        unordered_map<int,bool> map;
        int n=rooms.size();
        for(int i=0;i<n;i++){
            map[i]=false;
        }
        queue<int> q;
        q.push(0);
        while(!q.empty()){
            int k=q.size();
            while(k--){
                int a=q.front();
                q.pop();
                map[a]=true;
                for(int j=0;j<rooms[a].size() && rooms[a].size()!=0;j++){
                    if(map[rooms[a][j]]==false){
                        q.push(rooms[a][j]);    
                    }     
                }
            }
        }
        for(auto i:map){
            if(i.second==false){
                return false;
            }
        }
        return true;
    }

================================================
FILE: DSA Essentials Solutions/Hashing/ArrayIntersection.cpp
================================================
#include<bits/stdc++.h>
using namespace std;


vector<int> intersection(vector<int>& nums1, vector<int>& nums2) {
        
        unordered_map<int,int> map;
        vector<int> result;
        
        for(int i=0;i<nums1.size();i++)
        {
            map[nums1[i]]++;
        }      
        for(int i=0;i<nums2.size();i++)
        {
            if(map[nums2[i]] > 0)
            {
                result.push_back(nums2[i]);
                map[nums2[i]] = 0;
            }
        }    
        sort(result.begin(), result.end());
        return result;   
    }



================================================
FILE: DSA Essentials Solutions/Hashing/KSumSubarray.cpp
================================================
#include<bits/stdc++.h>
using namespace std;


int longestSubarrayKSum(vector<int> arr,int k){
    int n = arr.size();
	unordered_map<int,int> m;
	int pre = 0;

	int len = 0;

	for(int i=0;i<n;i++){
		pre += arr[i];

		if(pre==k){
			len = max(len,i+1);
		}

		if(m.find(pre-k)!=m.end()){
			len = max(len,i - m[pre-k]);
		}
		else{
			//store the first occ
			m[pre] = i;
		}

	}
	return len;

}


================================================
FILE: DSA Essentials Solutions/Heaps/MaximumProduct.cpp
================================================
#include<bits/stdc++.h>
using namespace std;


int maxProduct(vector<int>& nums) {
        priority_queue<int> q;
        for(int i=0; i<nums.size(); i++){
            q.push((nums[i]-1));
        }
        
        int p=q.top();
        q.pop();
        return p*q.top();
    }





================================================
FILE: DSA Essentials Solutions/Heaps/ReduceArraySizeToHalf.cpp
================================================

#include<bits/stdc++.h>
using namespace std;

int minSetSize(vector<int>& arr) {
        int n=arr.size();
        priority_queue<int> q;
        unordered_map<int,int> mp;
        for(int i=0; i<n; i++){
            mp[arr[i]]++;
        }
        for(auto pair:mp){
            q.push(pair.second);
        }
        int sum=0;
        int cnt=0;
        while(n-sum>n/2){
            sum+=q.top();
            q.pop();
            cnt++;
            
        }
        return cnt;
    }



================================================
FILE: DSA Essentials Solutions/Heaps/RelativeRanks.cpp
================================================
 #include<bits/stdc++.h>
using namespace std;
   vector<string> findRelativeRanks(vector<int>& score) {
        priority_queue<pair<int,int>> pq;
        for(int i=0; i<score.size(); i++){
            pq.push({score[i],i});
        }
       
    int n= score.size();
    
    vector<string> vec(n);
   
    
    int cnt=0;
    
    
    while(!pq.empty()){
        cnt++;
        
        if(cnt==1){
            cout<<"hey"<<endl;
            vec[pq.top().second].append("Gold Medal");
        }
        else if(cnt==2){
            vec[pq.top().second].append("Silver Medal");
        }
        else if(cnt==3){
            vec[pq.top().second].append("Bronze Medal");
        }
        else {
            vec[pq.top().second].append(to_string(cnt));
        }
        pq.pop();
    }
    return vec;
    }




================================================
FILE: DSA Essentials Solutions/Heaps/WeakestRows.cpp
================================================
#include<bits/stdc++.h>
using namespace std;


vector<int> kWeakestRows(vector<vector<int>>& mat, int k) {
        priority_queue <pair<int,int>, vector<pair<int,int>>, greater<pair<int,int>> > pq; //Min-heap
        for(int i=0;i<mat.size();i++)
        {
            int count=0;
            for(int j=0;j<mat[i].size();j++)
            {
                if(mat[i][j]==1)
                {
                    count++; //Counting the number of soldiers in each case
                }
            }
            pq.push(make_pair(count,i));
        }
        vector<int> x;
        while(k>0)
        {
            pair<int,int> temp=pq.top();
            x.push_back(temp.second);
            pq.pop();
            k--;
        }
        return x;
    }



================================================
FILE: DSA Essentials Solutions/Linked List/AlternateMerge.cpp
================================================
//Expected Time Complexity: O(n)

#include<bits/stdc++.h>
using namespace std;


class node{
public:
	int data;
	node* next;

	node(int data){
		this->data = data;
		next = NULL;
	}
};

node* apend(node* root, int d){
    if(root == NULL) return new node(d);
    node* temp = root;
    while(temp->next){
        temp = temp->next;
    }
    temp->next = new node(d);
    return root;
}

node* alternateMerge(node * root1, node* root2){
    //Complete this function 
    node* root = NULL;
    if(!root1) return root2;
    if(!root2) return root1;
    while(root1 && root2){
        root = apend(root, root1->data);
        root = apend(root, root2->data);
        root1 = root1->next;
        root2 = root2->next;
    }
    if(root1){
        while(root1){
            root = apend(root, root1->data);
            root1 = root1->next;
        }
    }
    if(root2){
        while(root2){
            root = apend(root, root2->data);
            root2 = root2->next;
        }
    }
    
    return root;
}


================================================
FILE: DSA Essentials Solutions/Linked List/BubbleSortOnLinkedList.cpp
================================================
//Expected Time Complexity: O(n^2)

#include <iostream>
using namespace std;
class node
{
public:
    int data;
    node *next;
    node(int data)
    {
        this->data = data;
        this->next = NULL;
    }
};
 
 
int len(node* head)
{
    node* temp = head ;
    int i = 0 ;
     while(temp!=NULL)
     {
         i++;
         temp=temp->next ;
     }
    
    return i ;
}
node* bubble_sort_LinkedList_itr(node* head)
{
    int n = len(head)-1;
  
   while(n--)
 
   {
       node* prev =NULL;
    node*cur = head;
    while(cur->next!=NULL)
    {
        if(cur->data >=cur->next->data)
        {
            
            if(prev==NULL)
            {
                //first node
                node* nxt = cur->next ;
                cur->next = nxt->next ;
                nxt->next = cur ;
               prev=nxt ;
                head = prev ;
               
                
            }
            
            else
            {
                
                node* nxt = cur->next ;
                prev->next = nxt ;
                cur->next = nxt->next ;
                nxt->next = cur ;
                prev = nxt ;
              
                
            }
            
        }
        else
        {
           
             prev = cur ; 
            cur=cur->next ;
           
        }
        
        
    
    }
       
   }
   
    
    
    return head ;
    
}
 


================================================
FILE: DSA Essentials Solutions/Linked List/DeleteTail.cpp
================================================
//Expected Time Complexity: O(n)

#include<bits/stdc++.h>
using namespace std;


class node{
public:
	int data;
	node* next;
    
	node(int data){
		this->data = data;
		next = NULL;
	}
};

node* deleteTail(node * head){
    //Complete this function 
     if (head == NULL)
        return NULL;
 
    if (head->next == NULL) {
        delete head;
        return NULL;
    }
 
    // Find the second last node
    node* second_last = head;
    while (second_last->next->next != NULL)
        second_last = second_last->next;
 
    // Delete last node
    delete (second_last->next);
 
    // Change next of second last
    second_last->next = NULL;
 
    return head;
}



================================================
FILE: DSA Essentials Solutions/Linked List/KthLastElement.cpp
================================================
//Expected Time Complexity: O(n)

#include<bits/stdc++.h>
using namespace std;


class node{
public:
	int data;
	node* next;

	node(int data){
		this->data = data;
		next = NULL;
	}
};

int kthLastElement(node * head,int k){
    //Complete this function to return kth last element
    node * fast = head;
    node * slow = head;
    
    int cnt = 0;
    while(cnt < k){
        fast = fast->next;
        cnt++;
    }
    
    while(fast!=NULL){
        slow = slow->next;
        fast = fast->next;
    }
    
    return slow->data;
}

================================================
FILE: DSA Essentials Solutions/Queues/FirstNonRepeatingLetter.cpp
================================================
//Expected Time Complexity: O(n)

#include <bits/stdc++.h>
using namespace std;
const int MAX_CHAR = 26;

vector<char> firstnonrepeating(vector<char> str)
{
	queue<char> q;
    vector<char> v;
	int charCount[MAX_CHAR] = { 0 };

	for (int i = 0; i<str.size(); i++) {

		q.push(str[i]);

		charCount[str[i] - 'a']++;

		while (!q.empty()) {
			if (charCount[q.front() - 'a'] > 1)
				q.pop();
			else {
				v.push_back(q.front());
				break;
			}
		}

		if (q.empty())
			v.push_back('0');
	}
	return v;
}



================================================
FILE: DSA Essentials Solutions/Queues/InterleaveTwoHalvesOfQueue.cpp
================================================
//Expected Time Complexity: O(n)

#include <bits/stdc++.h>
using namespace std;

queue<int> interLeave(queue<int> q){
    int n=q.size();
	queue<int> q1, q2; 
	for (int i=0;i<(n/2);i++) { 
		q1.push(q.front()); 
		q.pop(); //Expected Time Complexity: O(2^n)


	} 
	for (int i=0;i<(n/2);i++) { 
		q2.push(q.front()); 
		q.pop(); 
	} 

	for (int i=0;i<(n/2);i++) { 
		q.push(q1.front()); 
		q1.pop(); 
		q.push(q2.front()); 
		q2.pop(); 
	} 
    return q;
}

================================================
FILE: DSA Essentials Solutions/Queues/SortQueueWithConstantSpace.cpp
================================================
#include <bits/stdc++.h>
using namespace std;

int minIndex(queue<int> &q, int sortedIndex)
{
    int min_index = -1;
    int min_val = INT_MAX;
    int n = q.size();
    for (int i=0; i<n; i++)
    {
        int curr = q.front();
        q.pop();  
        
        if (curr <= min_val && i <= sortedIndex)
        {
            min_index = i;
            min_val = curr;
        }
        q.push(curr); 
    }
    return min_index;
}

void insertMinToRear(queue<int> &q, int min_index)
{
    int min_val;
    int n = q.size();
    for (int i = 0; i < n; i++)
    {
        int curr = q.front();
        q.pop();
        if (i != min_index)
            q.push(curr);
        else
            min_val = curr;
    }
    q.push(min_val);
}
  
void sortQueue(queue<int> &q)
{
    for (int i = 1; i <= q.size(); i++)
    {
        int min_index = minIndex(q, q.size() - i);
        insertMinToRear(q, min_index);
    }
}

queue<int> sortqueue(queue<int> &q)
{
    sortQueue(q);
    return q;
}
  

================================================
FILE: DSA Essentials Solutions/Recursion/2DArrayMerge.cpp
================================================
Hint: Divide, sort row-wise, sort col-wise

#include<bits/stdc++.h>
using namespace std;

void merge_row(vector<vector<int>> &mat,int i, int cs, int cm, int ce){
   vector<int> sorted;
   int x=cs;
   int y=cm+1;
   //cout<<x<<" "<<cm<<" "<<y<<" "<<ce<<endl;
   while(x<=cm && y<=ce){
       if(mat[i][x]<mat[i][y]){
           sorted.push_back(mat[i][x]);
           x++;
       }
       else{
           sorted.push_back(mat[i][y]);
           y++;
       }
   }
   
   
   while(x<=cm){
       sorted.push_back(mat[i][x]);
       x++;
   }
   while(y<=ce){
       sorted.push_back(mat[i][y]);
       y++;
   }
   int k=0;
   for(int j=cs; j<=ce; j++){
       mat[i][j]=sorted[k];
       k++;
   }
   return;
}
void merge_col(vector<vector<int>> &mat,int j, int rs, int rm, int re){
       vector<int> sorted;
   int x=rs;
   int y=rm+1;
   while(x<=rm && y<=re){
       if(mat[x][j]<mat[y][j]){
           sorted.push_back(mat[x][j]);
           x++;
       }
       else{
           sorted.push_back(mat[y][j]);
           y++;
       }
   }
   while(x<=rm){
       sorted.push_back(mat[x][j]);
       x++;
   }
   while(y<=re){
       sorted.push_back(mat[y][j]);
       y++;
   }
   int k=0;
   for(int i=rs; i<=re; i++){
       mat[i][j]=sorted[k];
       k++;
   }
   return;
}


void merge(int m, int n, vector<vector<int>> &mat, int rs, int rm, int re,int cs, int cm, int ce){
    
    
    
    //for sorting rows
    for(int i=rs; i<=re; i++){
        merge_row(mat,i,cs,cm,ce);
    }
    
    //for sorting columns
    for(int j=cs; j<=ce; j++){
        merge_col(mat,j,rs,rm,re);
    }
    return;

}

void merge_sort(int m, int n, vector<vector<int>> &mat, int rs, int re, int cs, int ce){
    //cout<<rs<<" "<<re<<endl;
    //cout<<cs<<" "<<ce<<endl;
     if(rs>=re && cs>=ce){
         return;
     }


     int rm=(rs+re)/2;
     int cm=(cs+ce)/2;
      
    // cout<<rs<<" "<<rm<<" "<<re<<" "<<cs<<" "<<cm<<" "<<ce<<endl; 
     
     
     //for dividing into subarrays
     merge_sort(m,n,mat,rs,rm,cs,cm);
     merge_sort(m,n,mat,rm+1,re,cs,cm);
     merge_sort(m,n,mat,rs,rm,cm+1,ce);
     merge_sort(m,n,mat,rm+1,re,cm+1,ce);

     
    //for merging sorted subarrays
     merge(m,n,mat,rs,rm,re,cs,cm,ce);
     return;
}

vector<vector<int>> mergeSort(int m, int n, vector<vector<int>> v){
    merge_sort(m,n,v,0,m-1,0,n-1);
    return v;
}


================================================
FILE: DSA Essentials Solutions/Recursion/AllOccurences.cpp
================================================
//Expected Time Complexity: O(N)


#include <bits/stdc++.h>
using namespace std;

vector<int> vec;


void helper(int k, vector<int> v, int i){
    if(i==v.size()){
        return;
    }
    if(v[i]==k){
        vec.push_back(i);
        
    }
    helper(k,v,i+1);
    return;
}

vector<int> findAllOccurences(int k, vector<int> v){
       vec.clear();
       helper(k,v,0);
       return vec;
}

================================================
FILE: DSA Essentials Solutions/Recursion/BinaryStrings.cpp
================================================
//Expected Time Complexity: O(2^n)

#include <bits/stdc++.h>
using namespace std;

vector<string> v;

void helper(string str,int n,int i){
    if(i==n){
        v.push_back(str);
        return;
    }
    string s1= str;
    s1.push_back('0');
    helper(s1,n,i+1);

    if(i>0 && str[i-1]=='0'){
       str.push_back('1');
       helper(str,n,i+1);
    }
    else if(i==0){
        str.push_back('1');
        helper(str,n,i+1);
    }

    return;
}

vector<string> binaryStrings(int n){
     v.clear();
     string str;
     helper(str,n,0);
     return v;

}

================================================
FILE: DSA Essentials Solutions/Recursion/FriendsParty.cpp
================================================
//Expected Time Complexity: O(2^n)

#include <iostream>
using namespace std;

int help(int n)
{
    if (n <= 0) return 0;
	if(n == 2 || n == 1) return n;
	return help(n - 1) + (n - 1) * help(n - 2);
}

int friendsPairing(int n){
    
    return help(n);
    
}

================================================
FILE: DSA Essentials Solutions/Recursion/PrintIncreasingNumbers.cpp
================================================
//Expected Time Complexity: O(n)

#include<bits/stdc++.h>
using namespace std;

void help(int i, int n, vector<int> &v)
{
    if(i > n) return;
    v.push_back(i);
    help(i+1, n, v);
}

vector<int> increasingNumbers(int N) {
    vector<int> v;
    help(1, N, v);
    return v;
}

================================================
FILE: DSA Essentials Solutions/Recursion/TilingProblem.cpp
================================================
//Expected Time Complexity: O(2^n)

#include <iostream>
using namespace std;

int tiles(int n,int m){
    if(n<m) return 1;
	int op1 = tiles(n-1, m);
	int op2 = tiles(n-m, m);
	return (op1 + op2);
}

int tillingProblem(int n, int m){
    return tiles(n, m);
}



================================================
FILE: DSA Essentials Solutions/Stacks/DuplicateParenthesis.cpp
================================================

//Expected Time Complexity: O(n)

#include<bits/stdc++.h>
using namespace std;

bool duplicateParentheses(string str){
    
    stack<char> Stack;
  
    for (char ch : str)
    {
        if (ch == ')')
        {
            char top = Stack.top();
            Stack.pop();
  
            int elementsInside = 0;
            while (top != '(')
            {
                elementsInside++;
                top = Stack.top();
                Stack.pop();
            }
            if(elementsInside < 1) {
                return true;
            }
        }
  
        else
            Stack.push(ch);
    }
  
    return false;
    
}


================================================
FILE: DSA Essentials Solutions/Stacks/MaximumRectangularAreaInHistogram.cpp
================================================
//Expected Time Complexity: O(n)

#include<bits/stdc++.h>
using namespace std;
 
int getMaxArea(vector<int> hist)
{
    int n = hist.size();
    stack<int> s;
 
    int max_area = 0; 
    int tp; 
    int area_with_top;
 
    int i = 0;
    while (i < n)
    {
        if (s.empty() || hist[s.top()] <= hist[i])
            s.push(i++);
        else
        {
            tp = s.top();  
            s.pop();  
            area_with_top = hist[tp] * (s.empty() ? i :
                                   i - s.top() - 1);
 
            if (max_area < area_with_top)
                max_area = area_with_top;
        }
    }
 
    while (s.empty() == false)
    {
        tp = s.top();
        s.pop();
        area_with_top = hist[tp] * (s.empty() ? i :
                                i - s.top() - 1);
 
        if (max_area < area_with_top)
            max_area = area_with_top;
    }
 
    return max_area;
}

================================================
FILE: DSA Essentials Solutions/Stacks/NextGreaterElement.cpp
================================================


#include<bits/stdc++.h>
using namespace std;

vector<int> nextGreaterElement(vector<int> arr){
    int n = arr.size();
    vector<int> arr1(n, 0);
    stack<int> s;
 
    for (int i = n - 1; i >= 0; i--)
    {
        while (!s.empty() && s.top() <= arr[i])
            s.pop();
 
        if (s.empty())
            arr1[i] = -1;        
        else
            arr1[i] = s.top();       
 
        s.push(arr[i]);
    }
        
    return arr1;
}
    


================================================
FILE: DSA Essentials Solutions/Stacks/ReverseANumberUsingStack.cpp
================================================
//Expected Time Complexity: O(n)

#include <bits/stdc++.h>
using namespace std;

int reverse(int n){
    
    int number = n;
    stack <int> st;
    while (number != 0) 
    {
        st.push(number % 10);
        number = number / 10;
    }
    
    int rev = 0;
    int i = 1;
      
    while (!st.empty()) 
    {
        rev = rev + (st.top() * i);
        st.pop();
        i = i * 10;
    }
      
    return rev;
	
}

================================================
FILE: DSA Essentials Solutions/Stacks/StockSpanProblem.cpp
================================================
//Expected Time Complexity: O(n)

#include <bits/stdc++.h>
using namespace std;

vector<int> stockSpanner(vector<int> &a){
	stack <int> s;
	int n = a.size();
	s.push(0);
	vector<int> arr(n, 1);
	for (int i = 1; i < n; i++) {
		while (!s.empty() and a[s.top()] <= a[i]) {
			s.pop();
		}
		if (!s.empty()) {
			arr[i] = i - s.top();
		} else arr[i] = i + 1;
		s.push(i);
	}
	return arr;
}

================================================
FILE: DSA Essentials Solutions/Strings/ArePermutation.cpp
================================================
//Expected Time Complexity= O(N log N)
//Hint: Permuatations are just different arrangements of same alphabets. Can you make the arrangement same?


#include <bits/stdc++.h>
using namespace std;
 
bool arePermutation(string str1, string str2)
{
    // Get lenghts of both strings
    int n1 = str1.length();
    int n2 = str2.length();
 
    // If length of both strings is not same,
    // then they cannot be Permutation
    if (n1 != n2)
      return false;
 
    // Sort both strings
    sort(str1.begin(), str1.end());
    sort(str2.begin(), str2.end());
 
    // Compare sorted strings
    for (int i = 0; i < n1;  i++)
       if (str1[i] != str2[i])
         return false;
 
    return true;
}

================================================
FILE: DSA Essentials Solutions/Strings/BinaryStringToNumber.cpp
================================================
Expected Time Complexity : O(N)


#include <iostream>
#include <string>
using namespace std;
 
// Function to convert binary to decimal
int binaryToDecimal(string n)
{
    string num = n;
    int dec_value = 0;
 
    // Initializing base value to 1, i.e 2^0
    int base = 1;
 
    int len = num.length();
    for (int i = len - 1; i >= 0; i--) {
        if (num[i] == '1')
            dec_value += base;
        base = base * 2;
    }
 
    return dec_value;
}
 

================================================
FILE: DSA Essentials Solutions/Strings/CheckPalindrome.cpp
================================================
Expected Time Complexity : O(N)


#include<bits/stdc++.h>
using namespace std;

bool isPalindrome(string str)
{
    // Start from leftmost and rightmost corners of str
    int l = 0;
    int h = str.length() - 1;
 
    // Keep comparing characters while they are same
    while (h > l)
    {
        if (str[l++] != str[h--])
        {
            return false;
        }
    }
    return true;
}

================================================
FILE: DSA Essentials Solutions/Strings/RemoveDuplicates.cpp
================================================
Expected Time Complexity : O(N)


#include <bits/stdc++.h>
using namespace std;

string removeDuplicate(string s){
    // your code goes here
    set<char> ss(s.begin(), s.end());
    string str;
 
    for (auto x : ss)
       str.push_back(x);
 
    return str;
}

================================================
FILE: DSA Essentials Solutions/Strings/StringCompression.cpp
================================================
Expected Time Complexity : O(N)



#include<bits/stdc++.h>
using namespace std;

int compress(vector<char>& chars) {
    
	int count_=1;
    string ans;
	
    for(int i=0;i<chars.size();i++)
    {
        while(i<chars.size()-1 && chars[i+1] == chars[i])
        {
            count_++;
            i++;
        }
        ans += chars[i];
        if(count_ == 1)
        {
            continue;
        }
        ans += to_string(count_);
        count_ = 1;
    }
    
     chars.clear();
    
     for(int i=0;i<ans.size();i++)
     {
          chars.push_back(ans[i]);
     } 
 
    return chars.size();
}

================================================
FILE: DSA Essentials Solutions/Strings/VowelFind.cpp
================================================
Expected Time Complexity : O(N)


#include<bits/stdc++.h>
using namespace std;

string vowel(string S){
    // your code goes here
    string out;
    for(auto x : S){
        if(x=='a' || x=='e' || x=='i' || x=='o' || x=='u') out.push_back(x);
    }
    return out;
} 

================================================
FILE: DSA Essentials Solutions/Trie/PrefixStrings.cpp
================================================
#include <bits/stdc++.h>
using namespace std;

class node{
    public:
    char ch;
    unordered_map<char,node*> next;
    bool isTerminal;
    node(char a){
        ch=a;
        bool isTerminal=false;
    }
};
class Trie{
    public:
    node*root= new node('\0');
    
    void insert(string str){

        node*temp=root;
        for(int i=0; i<str.length(); i++){
           if(temp->next.count(str[i])==0){
               temp->next[str[i]]=new node(str[i]);
           }
           temp=temp->next[str[i]];
        }
        temp->isTerminal=true;
        return;
    }
    void dfs(node*temp, vector<string> &v, string word ){
        if(temp->isTerminal){
            v.push_back(word);
        }
        if(temp->next.empty()){
            return;
        }
        for(auto p:temp->next){
          word.push_back(p.first);
          dfs(temp->next[p.first],v,word);
          word.pop_back();
        }
        return;
    }

    

    vector<string> find(string str){
        vector<string> v;
        node* temp=root;
        string word="";
        for(int i=0; i<str.length(); i++){
           if(temp->next.count(str[i])==0){
               return v;
           }
           word.push_back(str[i]);
           temp=temp->next[str[i]];
        }
        if(temp->isTerminal){
            v.push_back(word);
        }
        dfs(temp,v,word);
        sort(v.begin(),v.end());
        return v;
    }

    
    

};



vector<string> findPrefixStrings(vector<string> words, string prefix){
        Trie t;
        for(auto s:words){
            t.insert(s);
        }
        vector<string> res=t.find(prefix);
        return res;
    }


================================================
FILE: DSA Essentials Solutions/Vectors/MakeZeroes.cpp
================================================
#include<bits/stdc++.h>
using namespace std;

vector<vector<int>> makeZeroes(vector<vector<int>> arr){
    // your code goes here
    vector<int> r,c;
    int n = arr.size(), m = arr[0].size();
    
    for(int i=0; i<n; i++){
        for(int j=0; j<m; j++){
            if(arr[i][j] == 0){
                r.push_back(i); c.push_back(j);
            }
        }
    }
    
    for(auto x : r){
        for(int i=0; i<n; i++){
            arr[x][i] = 0;
        }
    }
    for(auto x : c){
        for(int i=0; i<n; i++){
            arr[i][x] = 0;
        }
    }
    
    return arr;
}

================================================
FILE: DSA Essentials Solutions/Vectors/RotateImage.cpp
================================================
#include<bits/stdc++.h>
using namespace std;

void rotate(vector<vector<int>>& matrix) {
        int n = matrix.size();
        int a = 0;
        int b = n-1;
        while(a<b){
            for(int i=0;i<(b-a);++i){
                swap(matrix[a][a+i], matrix[a+i][b]);
                swap(matrix[a][a+i], matrix[b][b-i]);
                swap(matrix[a][a+i], matrix[b-i][a]);
            }
            ++a;
            --b;
        }
    }


================================================
FILE: DSA Essentials Solutions/Vectors/SortFruits.cpp
================================================
#include<bits/stdc++.h>
using namespace std;

bool comp(pair<string,int> a, pair<string, int> b){
    return a.second < b.second;
}

vector<pair<string,int>> sortFruits(vector<pair<string,int>> v, string S){
    // your code  goes here
    if(S=="name") sort(v.begin(), v.end());
    else sort(v.begin(), v.end(), comp);
    return v;
}

================================================
FILE: DSA Essentials Solutions/Vectors/SortingCabs.cpp
================================================
#include<bits/stdc++.h>
using namespace std;

bool comp(pair<int,int> a, pair<int, int> b){
    float x = sqrt((a.first*a.first) + (a.second*a.second));
    float y = sqrt((b.first*b.first) + (b.second*b.second));
    return x < y;
}

vector<pair<int,int>> sortCabs(vector<pair<int,int>> v){
    // your code  goes here
    sort(v.begin(), v.end(), comp);
    return v;
}